The equation e^{sin x} - e^{-sin x} = 4 has . | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $e^{\sin x} = y$. Since $e^{\sin x} > 0$,we have $y > 0$. The equation becomes $y - \frac{1}{y} = 4$,which simplifies to $y^2 - 4y - 1 = 0$. S

Vedclass · Accepted Answer

no

Vedclass · Answer

(D) Let $e^{\sin x} = y$. Since $e^{\sin x} > 0$,we have $y > 0$. The equation becomes $y - \frac{1}{y} = 4$,which simplifies to $y^2 - 4y - 1 = 0$. Solving for $y$ using the quadratic formula,$y = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$. Since $y > 0$,we must have $y = 2 + \sqrt{5}$. Thus,$e^{\sin x} = 2 + \sqrt{5}$. Taking the natural logarithm on both sides,$\sin x = \ln(2 + \sqrt{5})$. Since $\sqrt{5} \approx 2.236$,$2 + \sqrt{5} \approx 4.236$. We know that $e \approx 2.718$,so $\ln(4.236) > \ln(e) = 1$. Therefore,$\sin x > 1$,which is impossible for any real $x$. Hence,the equation has no real solutions.

The equation $e^{\sin x} - e^{-\sin x} = 4$ has . . . . . . solutions.

Similar Questions

The number of values of $\theta$ in $[0, 2\pi]$ satisfying the equation $2\sin^2 \theta = 4 + 3\cos \theta$ is

The number of possible solutions of $\sin \theta + \sin 4 \theta + \sin 7 \theta = 0$ for $\theta \in (0, \pi)$ is:

If $(2\cos x - 1)(3 + 2\cos x) = 0$ for $0 \le x \le 2\pi$,then $x = $

The number of solutions of $\sin^{7} x + \cos^{7} x = 1$ for $x \in [0, 4\pi]$ is equal to:

If $1-\cos \theta = \sin \theta \cdot \sin \frac{\theta}{2}$,then the value of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module